Localization of a ring at one element

Question

Let $A$ be a commutative ring with identity, $f$ an element of $A$, let $g'=g/1$ be the image of the element $g$ of $A$ in $A_f$ under the natural homomorphism $A\rightarrow A_f$. The question is:

Is $A_{fg}$ isomorphic to $(A_f)_{g'}$?

Any comment will be appreciated!

Answer 1

A more general result holds:

Let $S,T$ be multiplicative sets in a commutative ring $A$, $i_S:A\to A_S$ the canonical homomorphism, and $T'=i_S(T)$. Then there is a unique isomorphism $j:A_{ST}\to (A_S)_{T'}$ such that $j\circ i_{ST}=i_{T'}\circ i_S$, that is, $j$ makes the following diagram to commute \begin{array} & & A\ \ \ \ \ \stackrel{i_S}\longrightarrow & A_S \\ & \downarrow i_{ST} &\downarrow i_{T'} \\ & A_{ST}\ \stackrel{j}\longrightarrow & (A_S)_{T'} \\ \end{array}

For proving this use the universality property of rings of fractions.